Proof:
First, we have
\(&I_2=\Re (\int _{-\pi }^{\pi }h(\alpha )\overline{L_2(\alpha )\partial _{\alpha }h(\alpha )}d\alpha )\\&=\int _{-\pi }^{\pi }\Re L_2(\alpha )(\Re h(\alpha )\Re h^{\prime }(\alpha )+\Im h(\alpha )\Im h^{\prime }(\alpha ))d\alpha \\&-\int _{-\pi }^{\pi }\Im L_2(\alpha )(\Re h(\alpha )\Im h^{\prime }(\alpha )-\Im h(\alpha )\Re h^{\prime }(\alpha ))d\alpha \\&=I_{2,1}+I_{2,s}.\) 

We could do the integration by parts to \(I_{2,1}\)  and have
\(|I_{2,1}|=|\frac{1}{2}\int _{-\pi }^{\pi }\frac{d\Re L_2(\alpha )}{d\alpha }((\Re h(\alpha ))^2+(\Im h(\alpha ))^2)d\alpha |\lesssim C(\Vert L_2(\alpha )\Vert _{C^1(\mathbb {T})})\Vert h(\alpha )\Vert _{L^2}^2.\) 

Moreover,
\(&I_1=\Re (\int _{-\pi }^{\pi }h(\alpha )\overline{L_1(\alpha )(\Lambda h)(\alpha )}d\alpha )\\&=\int _{-\pi }^{\pi }\Re L_1(\alpha )(\Re h(\alpha )\Lambda \Re h(\alpha )+\Im h(\alpha )\Lambda \Im h(\alpha ))d\alpha \\&-\int _{-\pi }^{\pi }\Im L_1(\alpha )(\Re h(\alpha )\Lambda \Im h(\alpha )-\Im h(\alpha )\Lambda \Re h(\alpha ))d\alpha \\&=I_{1,M}+I_{1,2}.\) 

We can still do the integration by parts to the \(I_{1,2}\)  and have
\(&\int _{-\pi }^{\pi }\Im L_1(\alpha )(\Re h(\alpha )\Lambda \Im h(\alpha )-\Im h(\alpha )\Lambda \Re h(\alpha ))d\alpha \\&=\int _{-\pi }^{\pi }(\Lambda (\Im L_1\Re h)(\alpha )-\Im L_1(\alpha )\Lambda \Re h(\alpha ))\Im h(\alpha )d\alpha .\) 

Moreover, for any \(g_1\in H^2(\mathbb {T})\) , \(g_2\in H^1(\mathbb {T})\) , we have
\(\Vert \Lambda (g_1g_2)-g_1\Lambda g_2\Vert _{L^2(\mathbb {T})}\lesssim \Vert g_2\Vert _{L^2(\mathbb {T})}\Vert g_1\Vert _{H^2(\mathbb {T})}.\) 

Hence
\(&I_{1,2}\lesssim \Vert \Im L_1(\alpha )\Vert _{H^2(\mathbb {T})}\Vert \Im h(\alpha )\Vert _{L^2(\mathbb {T})}\Vert \Re h(\alpha )\Vert _{L^2(\mathbb {T})}\\&\lesssim \Vert \Im L_1(\alpha )\Vert _{H^2(\mathbb {T})}\Vert  h(\alpha )\Vert _{L^2(\mathbb {T})}^2.\) 

Now we are left to control \(I_{1,M}+I_{2,s}\) . We have
\(&I_{1,M}+I_{2,s}=\int _{-\pi }^{\pi }\Re L_1(\alpha )(\Re h(\alpha )\Lambda \Re h(\alpha )+\Im h(\alpha )\Lambda \Im h(\alpha ))d\alpha \\&+\int _{-\pi }^{\pi }\Im L_2(\alpha )(\Re h(\alpha )\Im h^{\prime }(\alpha )-\Im h(\alpha )\Re h^{\prime }(\alpha ))d\alpha .\) 

Now we use a lemma from [1]}.
Lemma 5.4 
Let \(a\) , \(b\)  be real valued functions on \(\mathbb {T}\) , \(a(\alpha )\ge |b(\alpha )|\)  and satisfying \(a,b \in C^2(\mathbb {T})\) . Then we have
\(\Re \int _{\mathbb {T}}\overline{f(x)}(a(x)\Lambda f(x)+b(x)if^{\prime }(x))dx\ge -C(\Vert a\Vert _{C^2(\mathbb {T})}+\Vert b\Vert _{C^2(\mathbb {T})})\int _{\mathbb {T}}|f(x)|^2dx.\) 


Then from lemma REF , we have
\(I_{1,M}+I_{2,s}\lesssim C(\Vert L_1\Vert _{C^2(\mathbb {T})}+\Vert L_2\Vert _{C^2(\mathbb {T})})\Vert h(\alpha )\Vert _{L^2}^{2}.\) 
Then we get the result.
\(\Box \) 

Now let
\(&L_{z}^1(\alpha ,\gamma ,t)=-2\pi \lim _{\beta \rightarrow 0}(K(z(\alpha ,\gamma ,t)-z(\beta ,\gamma ,t),\tilde{f}(\alpha ,t)-\tilde{f}(\beta ,t))\tan (\frac{\alpha -\beta }{2}))\\\nonumber &=-2\pi \frac{\partial _{\alpha }z_1(\alpha ,\gamma ,t)+\partial _{\alpha }\tilde{f}_1(\alpha ,t)}{(\partial _{\alpha }z_1(\alpha ,\gamma ,t)+\partial _{\alpha }\tilde{f}_1(\alpha ,t))^2+(\partial _{\alpha }z_2(\alpha ,\gamma ,t)+\partial _{\alpha }\tilde{f}_2(\alpha ,t))^2},\) 

and
\(L_{z}^2(\alpha ,\gamma ,t)=(\frac{ic(\alpha )\gamma }{1+ic^{^{\prime }}(\alpha )\gamma t}+\frac{1}{1+ic^{\prime }(\alpha )\gamma t}p.v.\int _{-\pi }^{\pi } K(z(\alpha ,\gamma ,t)-z(\beta ,\gamma ,t),\tilde{f}(\alpha ,t)-\tilde{f}(\beta ,t))(1+ic^{\prime }(\beta )\gamma t)d\beta ).\) 

Since \({\rm supp}\,c(\alpha )\subset \lbrace \alpha | \lambda (\alpha )=1\rbrace \) , from (REF ), we have
\(&\frac{d}{dt}\Vert \partial _{\alpha }^5z(\alpha ,\gamma ,t)\Vert _{L^2_{\alpha }[-\pi ,\pi ]}^2\\&=\sum _{\mu =1,2}\underbrace{2\Re \int _{-\pi }^{\pi }\partial _{\alpha }^5z_{\mu }(\alpha )\overline{\lambda (\alpha )L_{z}^2(\alpha )\partial _{\alpha }^{6}z_{\mu }(\alpha )}d\alpha }_{\text{main term}}\underbrace{+2\Re \int _{-\pi }^{\pi }\partial _{\alpha }^5z_{\mu }(\alpha )\overline{\lambda (\alpha )L_{z}^1(\alpha ) \Lambda (\partial _{\alpha }^5z_{\mu })(\alpha )}d\alpha }_{\text{main term}}\\\nonumber &+B.T.\) 

From lemma (REF ), if \(-\Re L_{z}^1(\alpha )\ge |\Im L_z^2(\alpha )|\)  for \(\alpha \in [-2\delta ,2\delta ]\) , since \({\rm supp}\,\lambda \subset [-2\delta ,2\delta ]\) , we have
\(&\frac{d}{dt}\Vert \partial _{\alpha }^5z(\alpha ,\gamma ,t)\Vert _{L^2_{\alpha }[-\pi ,\pi ]}^2\lesssim B.T.\) 

Moreover, when \(t=0\) , \(\alpha \in [-2\delta ,2\delta ]\) , from (REF ), (REF ) and (REF ) we have
\(&-\Re L_z^1(\alpha ,\gamma ,0)=2\pi \frac{\partial _{\alpha }f_1^c(\alpha ,0)+\partial _{\alpha }\tilde{f}_1(\alpha ,0)}{(\partial _{\alpha }f_1^c(\alpha ,0)+\partial _{\alpha }\tilde{f}_1(\alpha ,0))^2+(\partial _{\alpha }f_2^c(\alpha ,0)+\partial _{\alpha }\tilde{f}_2(\alpha ,0))^2}\\&=2\pi \frac{\partial _{\alpha }f_1(\alpha ,0)}{(\partial _{\alpha }f_1(\alpha ,0))^2+(\partial _{\alpha }f_2(\alpha ,0))^2},\) 

and
\(&\Im L_z^2(\alpha ,0)=ic(\alpha )\gamma +\Im (p.v.\int _{-\pi }^{\pi } K(f^c(\alpha ,0)-f^c(\beta ,0),\tilde{f}(\alpha ,0)-\tilde{f}(\beta ,0))d\beta )\\&=ic(\alpha )\gamma .\) 

From (REF ), we could choose \(\delta _c\)  in (REF ) to be sufficiently small and have
\(\inf _{\alpha \in [-2\delta ,2\delta ]}(-\Re L_z^1(\alpha ,\gamma ,0)-|\Im L_z^2(\alpha ,\gamma ,0)|)>0.\) 

Then let
\(\Vert z\Vert _{RT}(t)=\sup _{\alpha \in [-2\delta ,2\delta ]}\frac{1}{|\Re L_z^1(\alpha ,\gamma ,t)+|\Im L_z^2(\alpha ,\gamma ,t)||}.\) 

If \(\Vert z\Vert _{RT}(t)< \infty \) , we have
\(\frac{d}{dt}\Vert z\Vert _{U}^2\lesssim C(\Vert z\Vert _{U}+\Vert z\Vert _{Arc}).\) 

Therefore, we could let \(\Vert z\Vert _{\tilde{U}}=\Vert z\Vert _{U}+\Vert z\Vert _{Arc}+\Vert z\Vert _{RT}\) . From (REF ), and the following lemma REF , we have
\(\frac{d}{dt}\Vert z\Vert _{\tilde{U}}^2\lesssim C(\Vert z\Vert _{\tilde{U}}).\) 

Then \(\Vert z(\alpha ,\gamma ,t)\Vert _{\tilde{U}}\)  is bounded for
sufficiently small time \(t_1\) . We claim that the bound and the time can be chosen such that it holds for all \(\gamma \in [-1,1]\) .
Lemma 5.5 
We have the following two estimates:
\(\frac{d}{dt}\Vert z\Vert _{Arc}\lesssim C(\Vert z\Vert _{\tilde{U}}),\) 

and
\(\frac{d}{dt}\Vert z\Vert _{RT}\lesssim C(\Vert z\Vert _{\tilde{U}}).\) 


